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  <h1 id="数学考研-线性代数" class="content-subhead">数学考研-线性代数</h1>
  <p>
    <span>1970-01-01</span>
    <span><span class="post-category post-category-math">Math</span></span>
    <span id="/public/article/数学考研-线性代数.html" class="leancloud_visitors" style="display:none" data-flag-title="数学考研-线性代数"></span>
  </p>
  <h1 id="_1">第二部分 线性代数</h1>
<p>3Blue1Brown 线性代数的本质 ：https://www.bilibili.com/video/BV1ys411472E</p>
<p><img class="pure-img" src="https://zromyk.gitee.io/myblog-figurebed/post/数学考研-线性代数.assets/截屏2020-12-19 20.50.08.jpg" alt="截屏2020-12-19 20.50.08" style="zoom:50%;" /></p>
<h2 id="1">第1讲 行列式的性质</h2>
<h4 id="1_1">1、行列式的几何意义</h4>
<p>几何意义：向量空间变换后的面积(2维)、体积(三维)变换比例。</p>
<h4 id="2">2、行列式基本公式</h4>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
      |A^T| &= |A| \\[1ex]
|\lambda A| &= \lambda^nA \\[1ex]
       |AB| &= |A||B| = |BA|
\end{split}\end{equation}
</script>
</p>
<h2 id="2_1">第2讲 矩阵</h2>
<h4 id="1_2">1、矩阵转置</h4>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
      (A^T)^T &= A \\[1ex]
      (A+B)^T &= A^T+B^T \\[1ex]
(\lambda A)^T &= \lambda A^T \\[1ex]
       (AB)^T &= B^TA^T
\end{split}\end{equation}
</script>
</p>
<h4 id="2_2">2、伴随矩阵</h4>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
 基本：AA^* &= A^*A = |A|E \\[1ex]
       A^* &= |A|A^{-1} ⟺ A^{-1} = \frac{1}{|A|}A^* \\[2em]
    (kA)^* &= k^{n-1}A^* \\[1ex]
   (A^T)^* &= (A^*)^T \\[1ex]
(A^{-1})^* &= (A^*)^{-1} = \frac{1}{|A|}A\\[1ex]
   (A^*)^* &= |A|^{n-2}A \\[2em]
     |A^*| &= |A|^{n-1} \\[1ex]
 |(A^*)^*| &= |A|^{(n-1)^2}
\end{split}\end{equation}
</script>
</p>
<h4 id="3">3、逆矩阵</h4>
<div class="pure-table-scrollable"><table class="pure-table pure-table-horizontal">
<thead>
<tr>
<th>奇异</th>
<th>可逆性</th>
</tr>
</thead>
<tbody>
<tr>
<td>奇异矩阵</td>
<td>可逆矩阵</td>
</tr>
<tr>
<td>非奇异矩阵</td>
<td>不可逆矩阵</td>
</tr>
</tbody>
</table></div>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
     基本：A^{-1} &= \cfrac{1}{|A|}A^* \\[1ex]
    基本：AA^{-1} &= A^{-1}A = E \\[1ex]
(\lambda A)^{-1} &= \frac{1}{\lambda}A^{-1} \\[1ex]
       (AB)^{-1} &= B^{-1}A^{-1} \\[2em]
   (A^{-1})^{-1} &= A \\[1ex]
      (A^T)^{-1} &= (A^{-1})^T
\end{split}\end{equation}
</script>
</p>
<div class="pure-table-scrollable"><table class="pure-table pure-table-horizontal">
<thead>
<tr>
<th>求逆矩阵的方法</th>
<th>公式</th>
</tr>
</thead>
<tbody>
<tr>
<td>1. 具体形式</td>
<td>
<script type="math/tex"> A^{-1}=\cfrac{E}{A} </script>
</td>
</tr>
<tr>
<td><strong>2. 初等变换</strong>（优先）</td>
<td>
<script type="math/tex"> [A｜E] = [E｜A]</script>
</td>
</tr>
<tr>
<td>3. 抽象形式</td>
<td>
<script type="math/tex"> AA^{-1}=A^{-1}A=E </script>
</td>
</tr>
</tbody>
</table></div>
<h4 id="4">4、矩阵的高次方</h4>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
  A &= P\Lambda P^{-1} \\[1ex]
A^n &= P\Lambda P^{-1}...P\Lambda P^{-1} = P\Lambda^nP^{-1} \\[2ex]
\Lambda &= \left[\begin{array}{} 
\lambda_1 \\
&\lambda_2 \\
&&\ddots \\
&&&\lambda_n
\end{array}\right]
\end{split}\end{equation}
</script>
</p>
<h2 id="3_1">第3讲 初等变换</h2>
<div class="pure-table-scrollable"><table class="pure-table pure-table-horizontal">
<thead>
<tr>
<th>设 <script type="math/tex"> A </script> 和 <script type="math/tex"> B </script> 为 <script type="math/tex"> m×n </script> 矩阵</th>
<th></th>
<th></th>
</tr>
</thead>
<tbody>
<tr>
<td>行变换</td>
<td>存在 <script type="math/tex"> m </script> 阶可逆矩阵 <script type="math/tex"> P </script>
</td>
<td>使得 <script type="math/tex"> PA=B </script>
</td>
</tr>
<tr>
<td>列变换</td>
<td>存在 <script type="math/tex"> n </script> 阶可逆矩阵 <script type="math/tex"> Q </script>
</td>
<td>使得 <script type="math/tex"> AQ=B </script>
</td>
</tr>
<tr>
<td>行列变换</td>
<td>存在 <script type="math/tex"> m </script> 阶可逆矩阵 <script type="math/tex"> P </script> 及 <script type="math/tex"> n </script> 阶可逆矩阵 <script type="math/tex"> Q </script>
</td>
<td>使得 <script type="math/tex"> PAQ=B </script>
</td>
</tr>
</tbody>
</table></div>
<p><details></p>
<p><summary><a style="color:blue">【2017数学一第21题】</a></summary></p>
<blockquote class="content-quote">
<p>例题：【2017数学一第21题】已知 <script type="math/tex">a</script> 是常数，且矩阵 <script type="math/tex; mode=display">\begin{equation}\begin{split}
A &= 
\left[\begin{array}{c c c}
1&2&a \\
1&3&0 \\
2&7&-a 
\end{array}\right]
\end{split}\end{equation}</script> 可经初等变换化为矩阵 <script type="math/tex; mode=display">\begin{equation}\begin{split}
B &= 
\left[\begin{array}{c c c}
1&a&2 \\
0&1&1 \\
-1&1&1 
\end{array}\right]
\end{split}\end{equation}</script>
</p>
<p>（I）求 <script type="math/tex">a</script>
</p>
<p>（II）求满足 <script type="math/tex">AP = B</script> 的可逆矩阵 <script type="math/tex">P</script>
</p>
<p>解：（II）</p>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
A &= 
\left[\begin{array}{c c c}
1&2&a \\
1&3&0 \\
2&7&-a 
\end{array}\right]
\sim
\left[\begin{array}{c c c}
1&2&a \\
0&1&-a \\
0&0&0 
\end{array}\right] \\[1em]
B &= 
\left[\begin{array}{c c c}
1&a&2 \\
0&1&1 \\
-1&1&1 
\end{array}\right]
\sim
\left[\begin{array}{c c c}
1&a&2 \\
0&1&1 \\
0&0&2-a 
\end{array}\right]
\end{split}\end{equation}
</script>
</p>
<p>由 <script type="math/tex">r(A) = r(B) = 2</script>，得 <script type="math/tex">a=2</script>
</p>
<p>（II）</p>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
A &= 
\left[\begin{array}{c c c}
1&2&2 \\
1&3&0 \\
2&7&-2 
\end{array}\right],
B &= 
\left[\begin{array}{c c c}
1&2&2 \\
0&1&1 \\
-1&1&1 
\end{array}\right]
\end{split}\end{equation}
</script>
</p>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
[A|B] &= 
\left[\begin{array}{c c c | c c c}
1&2&2&1&2&2 \\
1&3&0&0&1&1 \\
2&7&-2&-1&1&1  
\end{array}\right]
\sim
\left[\begin{array}{c c c | c c c}
1&0&6&3&4&4 \\
0&1&-2&-1&-1&-1 \\
0&0&0&0&0&0
\end{array}\right] \\[1em]
P &= 
\left[\begin{array}{c c c}
-6k_1+3&-6k_2+4&-6k_3+4 \\
2k_1-1&2k_2-1&2k_3-1 \\
k_1&k_2&k_3
\end{array}\right]
\sim
\left[\begin{array}{c c c}
1&1&1 \\
0&1&1 \\
0&0&k_3-k_2
\end{array}\right]
\end{split}\end{equation}
</script>
</p>
<p>由 <script type="math/tex">P</script> 可逆得到 <script type="math/tex">k_2 \ne k_3</script>，最后结果：</p>
<p>
<script type="math/tex; mode=display">
P = 
\left[\begin{array}{c c c}
-6k_1+3&-6k_2+4&-6k_3+4 \\
2k_1-1&2k_2-1&2k_3-1 \\
k_1&k_2&k_3
\end{array}\right], k_1,k_2,k_3\ 为任意常数，k_2 \ne k_3
</script>
</p>
<p>（若 <script type="math/tex">A</script> 和 <script type="math/tex">B</script> 可逆，可以通过 <script type="math/tex">P = A^{-1}B</script> 计算得到）</p>
</blockquote>
<p></details></p>
<h2 id="4_1">第4讲 矩阵的秩</h2>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
矩阵PQ可逆：&r(PAQ) = r(A) \\[2ex]
0 &\le r(A_{m*n}) \le \min\{m,n\} \\[2ex]
max\{r(A),r(B)\} &\le r(A,B) \le r(A) + r(B) \\[2ex]
r(AB) &\le r(A) + r(B) \\[2ex]
若A_{m*n}B_{n*l} = O&，则r(A) + r(B) \le n
\end{split}\end{equation}
</script>
</p>
<p>
<script type="math/tex">AB</script> 的列向量可以由 <script type="math/tex">A</script> 的列向量线性表示：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
C &= (c_0, c_1, \cdots, c_n) \\[1em]
AB &= 
\left(\begin{array}{c c c}
a_0 \\ 
a_1 \\ 
\vdots \\
a_n \\ 
\end{array}\right)
\left(\begin{array}{c c c}
b_0, b_1, \cdots, b_n 
\end{array}\right) \\[1em]
&=
\left(\begin{array}{c c c}
a_0b_0 & a_0b_1 & \cdots & a_0b_n \\ 
a_1b_0 & a_1b_1 & \cdots & a_1b_n \\ 
\vdots & \vdots & \ddots & \vdots \\
a_nb_0 & a_nb_1 & \cdots & a_nb_n \\ 
\end{array}\right) \\[2em]
于是，r(A,AB) &= r(A(E,B)) = r(A)
\end{split}\end{equation}
</script>
<br />
<script type="math/tex">BA</script> 的行向量可以由 <script type="math/tex">A</script> 的行向量线性表示：<br />
<script type="math/tex; mode=display">
r\left(\begin{array}{c c c}A \\ BA\end{array}\right) 
= r\left(\left(\begin{array}{c c c}E \\ B\end{array}\right)A\right)  = r(A)
</script>
</p>
<h2 id="5">第5讲 线性方程组</h2>
<h4 id="1_3">1、线性方程组的解性质</h4>
<div class="pure-table-scrollable"><table class="pure-table pure-table-horizontal">
<thead>
<tr>
<th>
<script type="math/tex"> Ax=0 </script>
</th>
<th>公式</th>
</tr>
</thead>
<tbody>
<tr>
<td>只有零解</td>
<td>
<script type="math/tex"> r(A) = n </script>
</td>
</tr>
<tr>
<td>有非零解</td>
<td>
<script type="math/tex"> r(A) \lt n </script>
</td>
</tr>
</tbody>
</table></div>
<div class="pure-table-scrollable"><table class="pure-table pure-table-horizontal">
<thead>
<tr>
<th>
<script type="math/tex"> Ax=b </script>
</th>
<th>公式</th>
</tr>
</thead>
<tbody>
<tr>
<td>无解</td>
<td>
<script type="math/tex"> r(A) \lt r(A,b) </script>
</td>
</tr>
<tr>
<td>有惟一解</td>
<td>
<script type="math/tex"> r(A) = r(A,b) = n </script>
</td>
</tr>
<tr>
<td>有无限多解</td>
<td>
<script type="math/tex"> r(A) = r(A,b) \lt n </script>
</td>
</tr>
</tbody>
</table></div>
<h4 id="2_3">2、线性方程组的解的结构</h4>
<ol>
<li>若 <script type="math/tex"> x=\xi_1,\ x=\xi_2 </script> 为 <script type="math/tex"> Ax=0 </script> 的解，则 <script type="math/tex"> x=\xi_1+\xi_2 </script> 也为其解</li>
<li>若 <script type="math/tex"> x=\xi </script> 为 <script type="math/tex"> Ax=0 </script> 的解，则 <script type="math/tex"> x=k\xi </script> 也为其解</li>
<li>若 <script type="math/tex"> r(A)=r </script> ，则 <script type="math/tex"> Ax=0 </script> 的解集的秩为 <script type="math/tex"> R_s=n-r=c </script> （c为最大无关组的个数）</li>
</ol>
<h4 id="3-a-ane0-ra-n">3、线性方程组的系数矩阵 <script type="math/tex">A</script> 的行列式 <script type="math/tex">|A|\ne0</script> ，秩 <script type="math/tex">r(A) = n</script>，则解唯一</h4>
<p>求解方法：</p>
<h5 id="1-axb-3">1. 克拉默法则 <script type="math/tex">Ax=b</script> ，以3阶举例</h5>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
x_1 =
\cfrac{
\left|\begin{array}{c c c}
b_1 & a_{12} & a_{13} \\
b_2 & a_{22} & a_{23} \\
b_3 & a_{32} & a_{33}
\end{array}\right|
}{
\left|\begin{array}{c c c}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|
},
x_2 =
\cfrac{
\left|\begin{array}{c c c}
a_{11} & b_1 & a_{13} \\
a_{21} & b_2 & a_{23} \\
a_{31} & b_3 & a_{33}
\end{array}\right|
}{
\left|\begin{array}{c c c}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|
},
x_3 =
\cfrac{
\left|\begin{array}{c c c}
a_{11} & a_{12} & b_1 \\
a_{21} & a_{22} & b_2 \\
a_{31} & a_{32} & b_3
\end{array}\right|
}{
\left|\begin{array}{c c c}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|
}
\end{split}\end{equation}
</script>
</p>
<h5 id="2-axb-xa-1axa-1b">2. 逆矩阵 <script type="math/tex">Ax=b</script>， 则 <script type="math/tex">x=A^{-1}Ax=A^{-1}b</script>
</h5>
<h2 id="6">第6讲 向量组</h2>
<h4 id="1_4">1、向量组</h4>
<p><strong>定义：</strong>在向量空间V的一组向量 <script type="math/tex"> A </script>： <script type="math/tex"> \alpha_1,\alpha_2,...,\alpha_m </script> ， 如果存在<strong>不全为零</strong>的数 <script type="math/tex"> k_1, k_2, ···, k_m </script> ，使 <script type="math/tex"> \alpha_1k_1, \alpha_2k_2, ···, \alpha_mk_m=0 </script> ，则称向量组 <script type="math/tex"> A </script> 是线性相关的，否则数 <script type="math/tex"> k_1, k_2, ···, k_m </script> 全为0时，称它是线性无关的。</p>
<div class="pure-table-scrollable"><table class="pure-table pure-table-horizontal">
<thead>
<tr>
<th>推论</th>
<th>线性相关</th>
<th>线性无关</th>
</tr>
</thead>
<tbody>
<tr>
<td></td>
<td>
<script type="math/tex"> r(A) \lt m </script>
</td>
<td>
<script type="math/tex"> r(A) = m </script>
</td>
</tr>
<tr>
<td>向量组 <script type="math/tex"> A </script> 为方阵</td>
<td>
<script type="math/tex"> ｜A｜ = 0 </script>
</td>
<td>
<script type="math/tex"> ｜A｜ \ne 0 </script>
</td>
</tr>
</tbody>
</table></div>
<p>若 <script type="math/tex"> R_A = r </script> ，则 <script type="math/tex"> r+1 </script> 个向量都线性无关，这 <script type="math/tex"> r </script> 向量 <script type="math/tex"> A_0 </script> 称为 <script type="math/tex"> A </script> 的一个最大无关组。</p>
<h4 id="2_4">2、向量组与向量组</h4>
<ol>
<li>向量组 <script type="math/tex"> B </script> 能由向量组 <script type="math/tex"> A </script> 线性表示（向量组 <script type="math/tex"> B </script> 中的每一个向量 <script type="math/tex"> b_i </script> 都能由向量组 <script type="math/tex"> A=(a_1,a_2,...,a_n) </script> 线性表示）， <script type="math/tex"> r(A) = r(A,B)或r(B) \le r(A) </script>
</li>
<li>向量组等价， <script type="math/tex"> r(A) = r(B) = r(A,B) </script>
</li>
</ol>
<h4 id="3_2">3、向量组等价（同维度，不要求向量个数相等）</h4>
<p>向量组 <script type="math/tex"> B </script> 能由向量组 <script type="math/tex"> A </script> 线性表示，向量组 <script type="math/tex"> A </script> 也能由向量组 <script type="math/tex"> B </script> 线性表示，则称向量组 <script type="math/tex"> A </script> ， <script type="math/tex"> B </script> 等价. ⟺  <script type="math/tex"> r(A)=r(B)=r(A,B) </script>
</p>
<h4 id="4_2">4、向量组在空间坐标中的应用</h4>
<h5 id="1_5">（1）基</h5>
<p>向量组 <script type="math/tex"> \alpha_1,\alpha_2,\alpha_3 </script> 为 <script type="math/tex"> R^3 </script> 的一个<strong>基</strong>，则 <script type="math/tex"> \alpha_1,\alpha_2,\alpha_3 </script>
<strong>线性无关</strong>：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
r(\alpha_1,\alpha_2,\alpha_3) &= 3 \\[2ex]
|(\alpha_1,\alpha_2,\alpha_3)| &= 0
\end{split}\end{equation}
</script>
</p>
<h5 id="2_5">（2）坐标中的向量表示</h5>
<p><strong>非零向量</strong> <script type="math/tex"> \xi </script> 在基 <script type="math/tex"> \alpha_1,\alpha_2,\alpha_3 </script> 下的表达式：<br />
<script type="math/tex; mode=display">
\xi = k_1\alpha_1 + k_2\alpha_2 + k_3\alpha_3
</script>
</p>
<h5 id="3_3">（3）过渡矩阵</h5>
<p>
<script type="math/tex; mode=display">
[\eta_1,\eta_2,\eta_3]C=[\xi_1,\xi_2,\xi_3]
</script>
</p>
<p>
<script type="math/tex"> C </script> 称为基 <script type="math/tex">[\eta_1,\eta_2,\eta_3]</script> 到 <script type="math/tex">[\xi_1,\xi_2,\xi_3]</script> 的过渡矩阵，实际是向量空间 <script type="math/tex">[\eta_1,\eta_2,\eta_3]</script> 通过矩阵 <script type="math/tex">C</script> 变换映射到另一个向量空间 <script type="math/tex">[\xi_1,\xi_2,\xi_3]</script> .</p>
<h2 id="7">第7讲 特征值和特征向量</h2>
<video style="border: 1px solid rgba(0, 0, 0, 1);" controls="controls" width="100%" src="/post/数学考研-线性代数.assets/特征值和特征向量.mov"></video>

<h4 id="1_6">1、特征值</h4>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
Ax&=\lambda x,\ x\neq0 \\[1ex]
|\lambda_0E-A|&=0 \\[1ex]
|A|&=\lambda_1\lambda_2...\lambda_n \\[1ex]
tr(A)&=\lambda_1+\lambda_2...+\lambda_n
\end{split}\end{equation}
</script>
</p>
<h4 id="2_6">2、特征向量</h4>
<p>几何意义：向量空间变换后方向不变的向量。</p>
<blockquote class="content-quote">
<p>3Blue1Brown 特征向量与特征值 https://www.bilibili.com/video/BV1ys411472E?p=14</p>
</blockquote>
<p>
<script type="math/tex; mode=display">
带入 \lambda_0，得[\lambda_0E-A]x=0，求特征向量
</script>
</p>
<h4 id="3_4">3、向量内积</h4>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
[x,y] &= x^Ty \\[1ex]
||x|| &= \sqrt{[x,x]} = \sqrt{x_1^2+x_2^2+...+x_n^2} \\[1ex]
0\le\theta&=\arccos\cfrac{[x,y]}{||x||*||y||}\le\pi \\[1ex]
[x,y] &= ||x||*||y||*\cos\theta
\end{split}\end{equation}
</script>
</p>
<p>正交： <script type="math/tex"> \{[x,y] = 0\} \Rightarrow \{向量x和y正交\} </script>
</p>
<p>正交变换：若 <script type="math/tex"> Q </script> 为正交矩阵，则线性变换 <script type="math/tex"> x=Qy </script> 称为<strong>正交变换</strong><br />
<script type="math/tex; mode=display">
||x||=\sqrt{x^Tx}=\sqrt{y^TQ^TQy}=||y||
</script>
</p>
<p><strong>在线性代数中，正交变换是线性变换的一种，它从实内积空间V映射到V自身，且保证变换前后内积不变。</strong></p>
<h4 id="3_5">3、施密特正交化</h4>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
\beta_1&=\alpha_1 \\
\beta_2&=\alpha_2-\cfrac{[\beta_1,\alpha_2]}{[\beta_1,\beta_1]}\beta_1 \\
\ ...\ &...\ ... \\
\beta_r&=\alpha_r-\cfrac{[\beta_1,\alpha_r]}{[\beta_1,\beta_1]}\beta_1-\cfrac{[\beta_2,\alpha_r]}{[\beta_2,\beta_2]}\beta_2-...-\cfrac{[\beta_{r-1},\alpha_r]}{[\beta_{r-1},\beta_{r-1}]}\beta_{r-1}
\end{split}\end{equation}
</script>
</p>
<p><img class="pure-img" src="https://zromyk.gitee.io/myblog-figurebed/post/数学考研-线性代数.assets/截屏2020-12-07 00.34.40.jpg" alt="截屏2020-12-07 00.34.40" style="zoom:50%;" /></p>
<h2 id="8">第8讲 相似理论</h2>
<h4 id="1_7">1、相似矩阵</h4>
<div class="pure-table-scrollable"><table class="pure-table pure-table-horizontal">
<thead>
<tr>
<th>关系</th>
<th>表达式</th>
<th></th>
<th>记号</th>
<th>秩</th>
<th>正负惯性指数</th>
<th>特征值</th>
<th>关系</th>
</tr>
</thead>
<tbody>
<tr>
<td>等价</td>
<td>
<script type="math/tex"> P^{-1}AQ=B_1 </script>
</td>
<td>可逆矩阵 <script type="math/tex"> P </script>，可逆矩阵 <script type="math/tex">Q</script>
</td>
<td>
<script type="math/tex"> A ≅ B_1 </script>
</td>
<td>相同</td>
<td></td>
<td></td>
<td>关系最弱</td>
</tr>
<tr>
<td>合同</td>
<td>
<script type="math/tex"> P^TAP=B_2 </script>
</td>
<td>可逆矩阵 <script type="math/tex"> P </script>
</td>
<td>
<script type="math/tex"> A ≃ B_2 </script>
</td>
<td>相同</td>
<td>相同</td>
<td></td>
<td></td>
</tr>
<tr>
<td>相似</td>
<td>
<script type="math/tex"> P^{-1}AP=B_3 </script>
</td>
<td>可逆矩阵 <script type="math/tex"> P </script>
</td>
<td>
<script type="math/tex"> A \sim B_3 </script>
</td>
<td>相同</td>
<td>相同</td>
<td>相同</td>
<td>关系最强</td>
</tr>
</tbody>
</table></div>
<blockquote class="content-quote">
<p>若 <script type="math/tex">Q</script> 为正交矩阵，则 <script type="math/tex">Q^TAQ=Q^{-1}AQ=B</script>
</p>
<p><strong>正(负)惯性指数：线性代数里矩阵的正(负)的特征值个数。</strong></p>
</blockquote>
<p>矩阵 <script type="math/tex"> A </script> 和 <script type="math/tex">B</script> 相似的必要条件：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
|A|&=|B| \\[1ex]
r(A)&=r(B)&矩阵的秩相等。\\[1ex]
tr(A)&=tr(B)&矩阵的迹相等，即正对角线上的元素之和相等。 \\[1ex]
\lambda_A&=\lambda_B&矩阵的特征值相等。\\[1ex]
|\lambda E - A|&=|\lambda E - B| \\[1ex]
r(\lambda E - A)&=r(\lambda E - B) \\[1ex]
n_i=n-r(\lambda_iE-A)&=n-r(\lambda_iE-B) \\[1em]
\lambda_i为n_i重根，n-r(\lambda_iE-A)也表示(\lambda_iE-A)x&=0的解中线性无关向量的个数。
\end{split}\end{equation}
</script>
</p>
<p>矩阵 <script type="math/tex"> A </script> 和 <script type="math/tex">B</script> 相似的性质：</p>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
(1)\ A\sim B &\Rightarrow A^T\sim B^T,\ A^{-1}\sim B^{-1},\ A^*\sim B^* \\[1ex]
             &\Rightarrow A^m\sim B^m,\ f(A)\sim f(B) \\[1em]
(2)\ A\sim B,\ B\sim C &\Rightarrow A\sim C \\[1em]
(3)\ A\sim B &\Rightarrow A+A^T \nsim B + B^T \\[1ex]
       &\Rightarrow A+A^{-1} \sim B + B^{-1}\ (2016考研数学一选择题5)
\end{split}\end{equation}
</script>
</p>
<h4 id="2_7">2、正交矩阵</h4>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
P为正交矩阵&⟺P^TP=E \\
&⟺PP^T=E \\
&⟺P^{-1}=P^T \\
&⟺P&为正交矩阵 \\
&⟺P^T&为正交矩阵 \\
&⟺P^{-1}&为正交矩阵 \\
&⟺P^*&为正交矩阵 \\
&⟺-P&为正交矩阵
\end{split}\end{equation}
</script>
</p>
<h4 id="3_6">3、相似对角化</h4>
<p><strong>定义：</strong>设 <script type="math/tex"> n </script> 阶矩阵 <script type="math/tex"> A </script> ，若存在 <script type="math/tex"> n </script> 阶可逆矩阵 <script type="math/tex"> P </script> 和正交矩阵 <script type="math/tex">Q</script> 使得， <script type="math/tex"> P^{-1}AP=Q^TAQ=\Lambda </script> ，其中 <script type="math/tex"> \Lambda </script> 是对角矩阵，则称可相似对角化，记 <script type="math/tex"> A\sim\Lambda </script> ，则称 <script type="math/tex"> \Lambda </script> 是 <script type="math/tex"> A </script> 的相似对角化.</p>
<p>（1）矩阵 <script type="math/tex">A</script> 的 <u>正交</u> 特征向量矩阵 <script type="math/tex">P</script> (<strong>是可逆矩阵，不一定是正交矩阵</strong>) 可以将矩阵 <script type="math/tex">A</script> 对角化：<br />
<script type="math/tex; mode=display">
\ P = (\xi_1,\xi_2,...,\xi_n) \\[2ex]
\xi_n为A的\underline{正交}特征向量(线性无关)，若特征向量不正交，需要施密特正交化
</script>
</p>
<p>
<script type="math/tex; mode=display">
P^{-1}AP=\Lambda=
\left[\begin{array}{c c c}
\lambda_1&& \\
&\ddots& \\
&&\lambda_n 
\end{array}\right]
</script>
</p>
<p>（2）矩阵 <script type="math/tex">A</script> 的 <u>单位化正交</u> 特征向量矩阵 <script type="math/tex">Q</script> (<strong>一定是正交矩阵</strong>) 可以将矩阵 <script type="math/tex">A</script> 对角化：<br />
<script type="math/tex; mode=display">
正交矩阵\ Q = (\eta_1,\eta_2,...,\eta_n) \\[2ex] 
\eta_n为\alpha_n的\underline{单位化正交}特征向量(线性无关)，若特征向量不正交，需要施密特正交化，再单位化
</script>
</p>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
Q^{T}AQ &= \Lambda =
\left[\begin{array}{c c c}
\lambda_1&& \\
&\ddots& \\
&&\lambda_n 
\end{array}\right]\\
同时Q^{-1}AQ &= \Lambda \\[2ex]
(特征向量&单位化之后依然为特征向量)
\end{split}\end{equation}
</script>
</p>
<h5 id="_2">充要条件</h5>
<ol>
<li>
<script type="math/tex"> A </script> 有 <script type="math/tex"> n </script> 个线性无关的特征向量⟺ <script type="math/tex"> A\sim\Lambda </script>
</li>
<li>
<script type="math/tex"> n_i=n-r(\lambda_iE-A) </script> ⟺ <script type="math/tex"> A\sim\Lambda </script>
</li>
</ol>
<blockquote class="content-quote">
<p>
<script type="math/tex"> n_i </script> 表示 <script type="math/tex"> \lambda_i </script> 为 <script type="math/tex"> n_i </script> 重根， <script type="math/tex"> n-r(\lambda_iE-A) </script> 表示 <script type="math/tex"> (\lambda_iE-A)x=0 </script> 的解中线性无关的向量个数。</p>
</blockquote>
<h5 id="_3">充分条件</h5>
<ol>
<li>
<script type="math/tex"> A </script> 是实对称矩阵 <script type="math/tex"> \Rightarrow A \sim \Lambda </script>
</li>
<li>
<script type="math/tex"> A </script> 有 <script type="math/tex"> n </script> 个互异特征值 <script type="math/tex"> \Rightarrow A \sim \Lambda </script>
</li>
<li>
<script type="math/tex"> A^2 = A \Rightarrow A \sim \Lambda </script>
</li>
<li>
<script type="math/tex"> A^2 = E \Rightarrow A \sim \Lambda </script>
</li>
</ol>
<h2 id="9">第9讲 二次型</h2>
<h4 id="1_8">1、二次型</h4>
<h5 id="lambda_1lambda_2lambda_3-neq-0">一、 <script type="math/tex"> \lambda_1、\lambda_2、\lambda_3 \neq 0 </script>
</h5>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
\cfrac{x^2}{a^2}(+)\cfrac{y^2}{b^2}(+)\cfrac{z^2}{c^2} &= 1\  (椭球面)   \\
\cfrac{x^2}{a^2}(+)\cfrac{y^2}{b^2}(-)\cfrac{z^2}{c^2} &= 1\ (单叶双曲面) \\
\cfrac{x^2}{a^2}(-)\cfrac{y^2}{b^2}(-)\cfrac{z^2}{c^2} &= 1\ (双叶双曲面) \\[1em]
\cfrac{x^2}{a^2}(+)\cfrac{y^2}{b^2}(-)\cfrac{z^2}{c^2} &= 0\  (二次锥面) \\
\end{split}\end{equation}
</script>
</p>
<h5 id="lambda_1lambda_2-neq-0-lambda_3-0">二、 <script type="math/tex"> \lambda_1、\lambda_2 \neq 0,\ \lambda_3 = 0 </script>
</h5>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
&\cfrac{x^2}{2p}(+)\cfrac{y^2}{2q} = z\ (椭球抛物面，pq同号) \ \ \ \ \ \ \ &\cfrac{x^2}{a^2}(+)\cfrac{y^2}{b^2} = 1\ (椭圆柱面)\\
&\cfrac{x^2}{2p}(-)\cfrac{y^2}{2q} = z\ (双曲抛物面，pq同号) \ \ \ \ \ \ \ &\cfrac{x^2}{a^2}(-)\cfrac{y^2}{b^2} = 1\ (双曲柱面)
\end{split}\end{equation}
</script>
</p>
<h5 id="lambda_1-neq-0-lambda_2lambda_3-0">三、 <script type="math/tex"> \lambda_1 \neq 0,\ \lambda_2、\lambda_3 = 0 </script>
</h5>
<p>
<script type="math/tex; mode=display">
\lambda_1 x^2 = by + cz + d\ (柱面)
</script>
</p>
<h4 id="2_8">2、正交变换</h4>
<p>正交变换：若 <script type="math/tex"> Q </script> 为正交矩阵，则线性变换 <script type="math/tex"> x=Qy </script> 称为<strong>正交变换</strong><br />
<script type="math/tex; mode=display">
||x||=\sqrt{x^Tx}=\sqrt{y^TQ^TQy}=||y||
</script>
</p>
<p>在线性代数中，正交变换是线性变换的一种，它从实内积空间V映射到V自身，且保证变换前后内积不变。</p>
<p>任意二次型，总有正交变换  <script type="math/tex"> x=Qy </script>  使之称为标准型：<br />
<script type="math/tex; mode=display">
x^TAx=y^T(Q^TAQ)y
</script>
</p>
<h4 id="3_7">3、二次型标准化方法</h4>
<h5 id="1_9">（1）正交相似变化法</h5>
<p>步骤：</p>
<ol>
<li>写出二次型对应的矩阵 <script type="math/tex"> A </script>
</li>
<li>求出 <script type="math/tex"> A </script> 的所有特征值： <script type="math/tex"> \lambda_1,\lambda_2,...,\lambda_n </script>
</li>
<li>求出对应于特征值的特征向量： <script type="math/tex"> \xi_1,\xi_2,...,\xi_n </script>
</li>
<li>将特征向量 <strong>施密特正交化</strong>，再 <strong>单位化</strong>，得： <script type="math/tex"> Q=(\eta_1,\eta_2,...,\eta_n) </script>
</li>
<li>作正交变换： <script type="math/tex"> x^TAx=y^T(Q^TAQ)y </script> ， <script type="math/tex"> Q^TAQ=\Lambda </script>
</li>
<li>得到标准形： <script type="math/tex"> f=\lambda_1y_1^2+\lambda_2y_2^2+...+\lambda_ny_n^2 </script>
</li>
</ol>
<h5 id="2_9">（2）拉格朗日配方法</h5>
<p>
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
f&=x_1^2+2x_2^2+5x_3^2+2x_1x_2+2x_1x_3+6x_2x_3 \\
 &=(x_1+x_2+x_3)^2+x_2^2+4x_3^2+4x_2x_3 \\
 &=(u_1)^2+u_2^2+4u_3^2+4u_2u_3 \\
 &=(u_1)^2+(u_2+2u_3)^2 \\
 &=(y_1)^2+(y_2)^2 \\[2ex]
 &=y_1^2+y_2^2
\end{split}\end{equation}
</script>
</p>
<p><details></p>
<p><summary><a style="color:blue">【2018数学一第20题】</a></summary></p>
<blockquote class="content-quote">
<p>例题：【2018数学一第20题】设实二次型 <script type="math/tex">f(x_1,x_2,x_3)=(x_1-x_2+x_3)^2+(x_2+x_3)^2+(x_1+ax_3)^2</script>，其中 <script type="math/tex">a</script> 是参数.</p>
<p>（I）求 <script type="math/tex">f(x_1,x_2,x_3) = 0</script> 的解；</p>
<p>（II）求 <script type="math/tex">f(x_1,x_2,x_3)</script> 的规范形；</p>
<p>解：（I）</p>
<p>
<script type="math/tex; mode=display">
\begin{cases}
&x_1&-&x_2&+&x_3 &= 0 \\
&&&x_2&+&x_3 &= 0 \\ 
&x_1&&&+a&x_3 &= 0
\end{cases} \\[3em]
A = 
\left[\begin{array}{r r r}
1&-1&1 \\
0&1&1 \\
1&0&a 
\end{array}\right]
\to
\left[\begin{array}{c c c}
1&0&2 \\
0&1&1 \\
0&0&a-2 
\end{array}\right]
</script>
</p>
<p>当 <script type="math/tex">a\ne2</script> 时 <script type="math/tex">r(A) = 3</script>，<script type="math/tex">f(x_1,x_2,x_3) = 0</script> 只有零解</p>
<p>当 <script type="math/tex">a=2</script> 时 <script type="math/tex">r(A) = 2</script>，<script type="math/tex">f(x_1,x_2,x_3) = 0</script> 有通解：</p>
<p>
<script type="math/tex; mode=display">
k\left[\begin{array}{r}
-2 \\
-1 \\
1 
\end{array}\right],k\in R
</script>
</p>
<p>（II）</p>
<p>当 <script type="math/tex">a\ne2</script> 时，令</p>
<p>
<script type="math/tex; mode=display">
\begin{cases}
y_1 &= x_1-x_2+x_3 \\
y_2 &= x_2+x_3 \\ 
y_3 &= x_1+ax_3
\end{cases} \\[3em]
\begin{equation}\begin{split}
f(x_1,x_2,x_3)
&=(x_1-x_2+x_3)^2+(x_2+x_3)^2+(x_1+ax_3)^2 \\
&=y_1^2+y_2^2+y_3^2
\end{split}\end{equation}
</script>
</p>
<p>得到 <script type="math/tex">f(x_1,x_2,x_3)</script> 的规范形为 <script type="math/tex">y_1^2+y_2^2+y_3^2</script>
</p>
<p>当 <script type="math/tex">a=2</script> 时，令<br />
<script type="math/tex; mode=display">
\begin{cases}
y_1 &= x_1-x_2+x_3 \\
y_2 &= x_2+x_3 \\ 
y_3 &= x_1+2x_3 = y_1+y_2
\end{cases} \\[3em]
\begin{equation}\begin{split}
f(x_1,x_2,x_3)
&=(x_1-x_2+x_3)^2+(x_2+x_3)^2+(x_1+ax_3)^2 \\
&=y_1^2+y_2^2+y_3^2 \\
&=y_1^2+y_2^2+(y_1+y_2)^2 \\
&=2y_1^2+2y_2^2+2y_1y_2 \\
&=2(y_1^2+\cfrac{1}{2}y_2)^2+\cfrac{3}{2}y_2^2 \\
&=2z_1^2+\cfrac{3}{2}z_2^2
\end{split}\end{equation}
</script>
<br />
得到 <script type="math/tex">f(x_1,x_2,x_3)</script> 的规范形为 <script type="math/tex">z_1^2+z_2^2</script>
</p>
</blockquote>
<p></details></p>
<h5 id="3_8">（3）已知二次型和标准化二次型，求正交矩阵</h5>
<p><details></p>
<p><summary><a style="color:blue">【2017数学一第21题】</a></summary></p>
<blockquote class="content-quote">
<p>例题：【2017数学一第21题】设二次型  <script type="math/tex"> f(x_1,x_2,x_3)=2x_1^2-x_2^2+ax_3^2+2x_1x_2-8x_1x_3+2x_2x_3 </script> ，在正交变换  <script type="math/tex"> x=Qy </script>  下的标准型为  <script type="math/tex"> \lambda_1y_1^2+\lambda_2y_2^2 </script> ，求  <script type="math/tex"> a </script>  的值以及一个正交矩阵  <script type="math/tex"> Q </script>
</p>
<p>1、二次型  <script type="math/tex"> f(x_1,x_2,x_3) </script>  对应的矩阵为：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
A &= 
\left[\begin{array}{r r r}
2&1&-4 \\
1&-1&1 \\
-4&1&a 
\end{array}\right]
\end{split}\end{equation}
</script>
</p>
<p>2、标准型  <script type="math/tex"> \lambda_1y_1^2+\lambda_2y_2^2 </script>  对应的对角矩阵为：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
\Lambda &= 
\left[\begin{array}{c c c}
\lambda_1&& \\
&\lambda_2& \\
&&0 
\end{array}\right]
\end{split}\end{equation}
</script>
<br />
3、即，表示矩阵  <script type="math/tex"> A </script>  的三个特征值为  <script type="math/tex"> \lambda_1,\lambda_2,0 </script>
</p>
<p>4、则，矩阵  <script type="math/tex"> A </script>  的行列式：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
|A| &= 
\left|\begin{array}{r r r}
2&1&-4 \\
1&-1&1 \\
-4&1&a 
\end{array}\right| = -3a+6=\lambda_1*\lambda_2*0=0 \\[1em]
a &= 2 
\end{split}\end{equation}
</script>
<br />
5、则得到完整的矩阵  <script type="math/tex"> A </script> ：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
A &= 
\left[\begin{array}{r r r}
2&1&-4 \\
1&-1&1 \\
-4&1&2
\end{array}\right]
\end{split}\end{equation}
</script>
<br />
6、求特征值  <script type="math/tex"> \lambda_1,\lambda_2,0 </script> ：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
|\lambda E - A| &= 
\left|\begin{array}{r r r}
\lambda-2&-1&4 \\
-1&\lambda+1&-1 \\
4&-1&\lambda-2 
\end{array}\right| = \lambda(\lambda+3)(\lambda-6)=0 \\[1em]
\lambda &= 6,-3,0
\end{split}\end{equation}
</script>
<br />
7、求特征值  <script type="math/tex"> 6,-3,0 </script>  对应的特征向量：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
[6E - A] &= 
\left[\begin{array}{r r r}
4&-1&4 \\
-1&7&-1 \\
4&-1&4
\end{array}\right] \to
\left[\begin{array}{r r r}
1&0&1 \\
0&1&0 \\
0&0&0
\end{array}\right] 
& \xi_1 = \left[\begin{array}{r r r}1 \\ 0 \\ -1\end{array}\right] \\[1em]
[-3E - A] &= 
\left[\begin{array}{r r r}
-5&-1&4 \\
-1&-2&-1 \\
4&-1&-5
\end{array}\right] \to
\left[\begin{array}{r r r}
1&1&0 \\
0&1&1 \\
0&0&0
\end{array}\right] 
& \xi_2 = \left[\begin{array}{r r r}1 \\ -1 \\ 1\end{array}\right]\\[1em]
[- A] &= 
\left[\begin{array}{r r r}
-2&-1&4 \\
-1&1&-1 \\
4&-1&-2 
\end{array}\right] \to
\left[\begin{array}{r r r}
1&0&-1 \\
0&1&-2 \\
0&0&0
\end{array}\right] 
& \xi_3 = \left[\begin{array}{r r r}1 \\ 2 \\ 1\end{array}\right]\\[1em]
\end{split}\end{equation}
</script>
<br />
8、向量  <script type="math/tex"> \xi_1,\xi_2,\xi_3 </script>  彼此正交，可以直接单位化得  <script type="math/tex"> \eta_1,\eta_2,\eta_3 </script>
</p>
<p>[<strong>9、如果  <script type="math/tex"> \xi_1,\xi_2,\xi_3 </script>  不正交，需要用施密特正交化成正交向量，然后再进行单位化</strong>]</p>
<p>10、相似对角化的两种方法：<br />
<script type="math/tex; mode=display">
\begin{equation}\begin{split}
存在可逆矩阵 P &= (\xi_1,\xi_2,\xi_3)，使得P^{-1}AP &= \Lambda \\[1em]
存在正交矩阵 Q &= (\eta_1,\eta_2,\eta_3)，使得Q^TAQ &= \Lambda 
\end{split}\end{equation}
</script>
</p>
<p>
<script type="math/tex; mode=display">
Q = (\eta_1,\eta_2,\eta_3) 可以由 P = (\xi_1,\xi_2,\xi_3) 的列向量单位化得到
</script>
</p>
</blockquote>
<p></details></p>
<div class="pure-table-scrollable"><table class="pure-table pure-table-horizontal">
<thead>
<tr>
<th>矩阵的一系列求解</th>
<th>公式</th>
<th>应用</th>
</tr>
</thead>
<tbody>
<tr>
<td>矩阵</td>
<td>
<script type="math/tex; mode=display">A = \left[\begin{array}{r r} 0&1 \\ 1&1 \end{array}\right]</script>
</td>
<td></td>
</tr>
<tr>
<td>特征值</td>
<td>
<script type="math/tex; mode=display">\lambda_1=\cfrac{1-\sqrt{5}}{2},\lambda_2=\cfrac{1+\sqrt{5}}{2}</script>
</td>
<td></td>
</tr>
<tr>
<td><strong>正交</strong> 特征向量（若不正交需要施密特正交化）</td>
<td>
<script type="math/tex; mode=display">\xi_1 = \left[\begin{array}{c c} 2 \\ 1+\sqrt{5} \end{array}\right],\xi_2 = \left[\begin{array}{c c} 2 \\ 1-\sqrt{5} \end{array}\right]</script>
</td>
<td></td>
</tr>
<tr>
<td>可逆 <strong>正交</strong> 特征向量矩阵</td>
<td>
<script type="math/tex; mode=display">P = \left[\begin{array}{r r} 2.0000&2.0000 \\ 3.2360&-1.2360 \end{array}\right]</script>
</td>
<td></td>
</tr>
<tr>
<td>相似对角化</td>
<td>
<script type="math/tex; mode=display">P^{-1}AP = \Lambda = \left[\begin{array}{r r} 1.6180 & 0.0000 \\ 0.0000 & -0.6180 \end{array}\right]</script>
</td>
<td>
<script type="math/tex; mode=display">A^n = P\Lambda^nP^{-1}</script>
</td>
</tr>
<tr>
<td>可逆单位化正交特征向量矩阵</td>
<td>
<script type="math/tex; mode=display">Q = P单位化=\left[\begin{array}{r r} 0.5257 & 0.8507 \\ 0.8507 & -0.5257 \end{array}\right]</script>
</td>
<td></td>
</tr>
<tr>
<td>相似对角化</td>
<td>
<script type="math/tex; mode=display">Q^TAQ = \Lambda = \left[\begin{array}{r r} 1.6180 & 0.0000 \\ 0.0000 & -0.6180 \end{array}\right]</script>
</td>
<td>
<script type="math/tex; mode=display">A^n = Q\Lambda^nQ^T</script>
</td>
</tr>
</tbody>
</table></div>
</div>
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